An NMOS differential amplifier employing equal drain resistors, RD=47 kΩ, has a differential gai

(a) The differential gain $A_{d}$ of an NMOS differential amplifier is given by $A_{d} = g_{m}R_

An NMOS differential amplifier employing equal drain...

An NMOS differential amplifier employing equal drain resistors, $R_{D}=47 \mathrm{k} \Omega,$ has a differential gain $A_{d}$ of $20 \mathrm{V} / \mathrm{V}$ (a) What is the value of $g_{m}$ for each of the two transistors? (b) If each of the two transistors is operating at an overdrive voltage $V_{O V}=0.2 \mathrm{V}$, what must the value of $1 \mathrm{be}$ ? (c) For $v_{i d}=0,$ what is the de voltage across each $R_{D} ?$ (d) If $v_{i d}$ is 20 -m $V$ peak-to-peak sine wave applied in a balanced manner but superimposed on $V_{C M}=0.5 \mathrm{V}$, what is the lowest value that $V_{D D}$ must have to ensure saturation mode operation for $Q_{1}$ and $Q_{2}$ at all times? Assume $V_{t}=0.5 \mathrm{V}$

Transcript

00:02 So in this question, we have an n -mos differential amplifier, right? so we have an m -m -m -m -m -m differential amplifier.

00:23 And this amplifier is employing equal drain resistors, right? so we have equal -drain resistors.

00:37 Now there are five powers that we are required to solve.

00:46 And the information that is when given to us is that the rd is equal to 47 kiloohom.

00:58 And it has a differential gain of a differential gain that is a sub d.

01:14 Of 20 volts per volt right now what we are required to do over here is that first of all we will solve part a right so in part a what is required by us to do let's take a look at that so in part a the question states that what is the value of m for each of the two transistors, right? so let's take a look at the solution for part a.

01:52 For part a, the solution is very simple, right? so first of all, what we are going to do is that, that recalls your previous knowledge of the expression for the differential gain, right? so the expression for the differential gain, absolute value of differential gain is equal to the product of g sub m into r sub d.

02:22 Right.

02:23 Now as you can see over here, we are given the value of the differential gain as well as resistance, right? so it means that we can use this equation to find out the value of trans conductors.

02:39 Gm right g sub m so it means that we need to um rearrange this equation right so let's rearrange it and when we'll rearrange it we'll get that it is equal to g sub m it's equal to a sub d divided by r d right so we have the value of a sub d and r sub d so just plug it over here and we'll get that it is equal to 20 divided by 47 into 10 raised to bar 3 item you solve it you'll get that it is equal to 425 .5 .5 .53 micro ampere per voltage right so this is your solution for part a right so transconductance is equal to 425 .5 .53 microse ampere per volt right now let's move forward to part b so in part b what we are required to do is that in part b the question states that if each of the two transistors they are operating at an overdrive voltage right overdrive voltage that is v sub o v and it's equal to 0 .2 volt right so we need to find out the value of i right now that's the grant so let's take a look at the solution of the solution for part b is also very simple now what we are going to do is that we are going to use the value of the trance conductance right transconductance gm right g sub m and the overdrive voltage that is this given voltage over drive voltage right and that is how we are going to determine the value of the bias current right so bias grant i so for this we are going to use transconductance and the overdrive voltage.

05:17 So we are going to choose the expression for the trans conductance that is known to us.

05:23 It's two times the i subd divided by v sub ov, right? so we are going to rearrange it for the current, right? and that would be j sub m into v sub ov divided by two.

05:45 Right so just simply plug in the values for the trans conductance and the overdrive voltage right and trans conductance is 425 .35 into 10 x to 1 negative 6 and overdrive voltage is equal to 0 .2 right so just calculated and we'll get that the value of i is equal to 85 .1 .1 micro -ampure.

06:30 Right.

06:31 So this is the solution for part b.

06:34 Now let's take a look at part c...

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